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Math Help - Integral question and answer confirmation please!

  1. #1
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    Integral question and answer confirmation please!

    I am suppose to evaluate the indefinite integral of
    dx / square root of (1 - x^2 +2x)..
    I first broke down 1-2^x + 2x into 2-(x-1)^2
    I then took the antiderivative of dx/(2-(x-1)^2) and came up with the answer of [sin^-1(x-1) / 2] + C
    Can someone please tell me if this is correct, or what i can do to make it right? I took the derivavtive and it seems i might have made a small error somewhere in finding the antiderivative but im not soo sure where...


    Also, if someone can help me solve this question, thatd be great, thanks!!
    Compute the integral from -1 ->3 of the square root of (6-2x^2 +4x)dx .
    I can't find the antiderivative of this function...and substitution doesnt seem to work...
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  2. #2
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    Quote Originally Posted by shmayer View Post
    I am suppose to evaluate the indefinite integral of
    dx / square root of (1 - x^2 +2x)..
    I first broke down 1-2^x + 2x into 2-(x-1)^2
    I then took the antiderivative of dx/(2-(x-1)^2) and came up with the answer of [sin^-1(x-1) / 2] + C
    Can someone please tell me if this is correct, or what i can do to make it right? I took the derivavtive and it seems i might have made a small error somewhere in finding the antiderivative but im not soo sure where...


    Compute the integral from -1 ->3 of the square root of (6-2x^2 +4x)dx .
    I can't find the antiderivative of this function...and substitution doesnt seem to work...
    I thought I saw these two exact problems before ...

    http://www.mathhelpforum.com/math-he...gral-help.html

    ... did you even bother to look?
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  3. #3
    Member Abu-Khalil's Avatar
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    \int\frac{dx}{\sqrt{1-x^2+2x}}=\int\frac{dx}{\sqrt{2-(x-1)^2}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{1-\left(\frac{x-1}{\sqrt{2}}\right)^2}}=\arcsin \left(\frac{x-1}{\sqrt{2}}\right)+C

    PD: Ups. Late post.
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